Given triangle as given in the below image, assume that angles $BCA, CA'B$ and $AB'C$ have equal measure.
Show that $(AC)^2 + (BC)^2 = AB(AB'+ A'B)$. What additional assumption
gives the Pythagorean Theorem?
Need first show that if angles $BCA, CA'B$ and $AB'C$ have equal measure, show that $(AC)^2 + (BC)^2 = AB(AB'+ A'B)$.
Consider the $3$ triangles: $BCA, CA'B, AB'C$, with orientation shown in the same cck order, and the common angle in center, with base being respectively $AB, BC, CA$.
It is obvious that only similarity can be worked out here.
Edit : Kindly see my first comment below for doubts about whether only similarity can be worked out here.
*Edit 2: * Using the @GCab hint: triangles $BCA, CA'B$ common angles: $C=A', B=B$ (but, now the 2nd triangle would be listed as $BA'C$ to have the same cck orientation of angles and sides). Unable to find any (need one similar side between common angles, to apply similarity) common sides. Anyway the third angle of both triangles have to be equal to each other. Similarly, for triangles $BCA, CB'A$. Can use two ratios for these two pairs of slr. triangles (i.e., (i)$BCA, BA'C$. , (ii) $BCA, CB'A$) to get : $\frac{AB}{CB} = \frac{CB}{A'B}$, $\frac{AB}{AC} = \frac{AC}{AB'}$. This would lead to two equalities among products to be summed up, to solve the first part easily.
But, what is the intuition gained?
The intuition I expected was in algebraic terms, and not that for second question to be solved $A'=B'$, i.e. there can be only one position on $AB$ that can raise at $90^0$ to meet vertex $A$, even if I am correct.
The triangles $ACB$, $AB'C$ and $BA'C$ are similar, having two angles (and one edge) in common.
Thus establish the similarity proportionalities $$ \left\{ \begin{gathered} \frac{{c_{\,a} }} {b} = \frac{d} {a} = \frac{b} {c} \hfill \\ \frac{{c_{\,b} }} {a} = \frac{d} {b} = \frac{a} {c} \hfill \\ \end{gathered} \right. $$ which then can be developed as follows $$ \eqalign{ & \left\{ \matrix{ ac_{\,a} = bd \hfill \cr bc_{\,b} = ad \hfill \cr dc = ab \hfill \cr} \right. \cr & \left\{ \matrix{ cac_{\,a} = bab \hfill \cr cbc_{\,b} = aab \hfill \cr dc = ab \hfill \cr} \right. \cr & \left\{ \matrix{ cc_{\,a} = b^{\,2} \hfill \cr cc_{\,b} = a^{\,2} \hfill \cr dc = ab \hfill \cr} \right. \cr & c\left( {c_{\,a} + c_{\,b} } \right) = a^{\,2} + b^{\,2} \cr} $$ to obtain the starting relation in your post.
Then to obtain that $c_a+c_b=c$ you shall clearly have that $\gamma = \pi/2$.